The equation of a circle $C$ is $x^2+y^2-4x+12y+36 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-4x) + (y^2+12y) = -36$ $(x^2-4x+4) + (y^2+12y+36) = -36 + 4 + 36$ $(x-2)^{2} + (y+6)^{2} = 4 = 2^2$ Thus, $(h, k) = (2, -6)$ and $r = 2$.